In a hydrogen atom the electron makes a transition from `(n+1)^(th)` level to the`n^(th)` level .If ` n gt gt 1 ` the frequency of radiation emitted is proportional to :

Options

A. `1/(n^(2))`
B. `1/n`
C. `1/(n^(3))`
D. `1/(n^(4))`

Explanation

To solve the problem, we need to determine how the frequency of radiation emitted during a transition from the \((n+1)^{th}\) level to the \(n^{th}\) level in a hydrogen atom is related to \(n\) when \(n\) is very large.

### Step-by-Step Solution:

1. **Energy Levels in Hydrogen Atom**:
The energy of an electron in a hydrogen atom at level \(n\) is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
Therefore, the energy at the \((n+1)^{th}\) level is:
\[ E_{n+1} = -\frac{13.6 \, \text{eV}}{(n+1)^2} \]
and the energy at the \(n^{th}\) level is:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

2. **Energy Change During Transition**:
The change in energy (\(\Delta E\)) when the electron transitions from the \((n+1)^{th}\) level to the \(n^{th}\) level is:
\[ \Delta E = E_n - E_{n+1} \]
Substituting the energy values:
\[ \Delta E = \left(-\frac{13.6}{n^2}\right) - \left(-\frac{13.6}{(n+1)^2}\right) \]
Simplifying this gives:
\[ \Delta E = -\frac{13.6}{n^2} + \frac{13.6}{(n+1)^2} \]

3. **Finding a Common Denominator**:
To combine the two fractions, we can find a common denominator:
\[ \Delta E = 13.6 \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} \right) \]
The expression \((n+1)^2 - n^2\) simplifies to:
\[ (n^2 + 2n + 1) - n^2 = 2n + 1 \]
Therefore:
\[ \Delta E = 13.6 \cdot \frac{2n + 1}{n^2(n+1)^2} \]

4. **Approximation for Large \(n\)**:
Since \(n \gg 1\), we can approximate \(2n + 1 \approx 2n\) and \((n+1)^2 \approx n^2\). Thus:
\[ \Delta E \approx 13.6 \cdot \frac{2n}{n^2 \cdot n^2} = 13.6 \cdot \frac{2}{n^3} \]

5. **Relating Energy to Frequency**:
The energy of the emitted photon is related to its frequency (\(f\)) by Planck's equation:
\[ \Delta E = h f \]
Therefore:
\[ f = \frac{\Delta E}{h} \approx \frac{13.6 \cdot 2}{h \cdot n^3} \]
This shows that the frequency \(f\) is inversely proportional to \(n^3\):
\[ f \propto \frac{1}{n^3} \]

### Final Result:
The frequency of radiation emitted is proportional to:
\[ \frac{1}{n^3} \]